A bowling ball is traveling at 7.6 meters per second when it hits a pin. If the bowling ball has a mass of 6 kilograms, how much kinetic energy does it have when it hits the pin?

KE=12mv2

(1 point)
Responses

22.8 J
22.8 J

346.65 J
346.65 J

693.12 J
693.12 J

173.28 J

1 answer

To find the kinetic energy (KE) of the bowling ball, we can use the formula:

\[ KE = \frac{1}{2}mv^2 \]

Where:

  • \(m\) is the mass of the object in kilograms (6 kg for the bowling ball),
  • \(v\) is the velocity in meters per second (7.6 m/s).

Plugging in the values:

\[ KE = \frac{1}{2} \times 6 , \text{kg} \times (7.6 , \text{m/s})^2 \]

First, we calculate \( (7.6 , \text{m/s})^2 \):

\[ (7.6)^2 = 57.76 , \text{m}^2/\text{s}^2 \]

Now substitute back into the equation:

\[ KE = \frac{1}{2} \times 6 \times 57.76 \]

Calculating further:

\[ KE = 3 \times 57.76 = 173.28 , \text{J} \]

So, the kinetic energy of the bowling ball when it hits the pin is 173.28 J.